04. Partial Differentiation¶

Boas Chapter 4 — Partial differentiation is a core tool across all physical sciences, including thermodynamics, electromagnetism, and fluid dynamics, because almost every law of physics is described by multivariable functions.

Learning Objectives¶

Upon completing this lesson, you will be able to:

Understand the definition and notation of partial derivatives and compute partial derivatives of multivariable functions
Apply the multivariable chain rule to find derivatives of composite functions
Perform implicit differentiation and apply it to physical relationships
Classify extrema and saddle points using the second derivative test
Solve constrained optimization problems using the Lagrange multiplier method
Understand the conditions for exact differentials and derive Maxwell relations in thermodynamics
Expand multivariate Taylor series and apply them to physics approximations

1. Basics of Partial Differentiation¶

1.1 Multivariable Functions and Partial Derivatives¶

The partial derivative of a multivariable function $f(x, y, z, \ldots)$ with respect to a specific variable is obtained by treating all other variables as constants and differentiating only with respect to that variable:

$$ \frac{\partial f}{\partial x} = \lim_{\Delta x \to 0} \frac{f(x + \Delta x, y, z, \ldots) - f(x, y, z, \ldots)}{\Delta x} $$

Notation: $\frac{\partial f}{\partial x}$, $f_x$, $\partial_x f$

Higher-order partial derivatives:

$$ \frac{\partial^2 f}{\partial x^2}, \quad \frac{\partial^2 f}{\partial x \partial y}, \quad \frac{\partial^2 f}{\partial y \partial x} $$

Schwarz's theorem: If the second partial derivatives of $f$ are continuous, the order of differentiation can be exchanged: $$\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}$$

import numpy as np
import sympy as sp

# SymPy를 이용한 편미분
x, y, z = sp.symbols('x y z')
f = x**2 * y + sp.sin(x * y) + z * sp.exp(x)

print(f"f = {f}")
print(f"∂f/∂x = {sp.diff(f, x)}")
print(f"∂f/∂y = {sp.diff(f, y)}")
print(f"∂f/∂z = {sp.diff(f, z)}")

# 2차 편미분
print(f"\n∂²f/∂x² = {sp.diff(f, x, 2)}")
print(f"∂²f/∂x∂y = {sp.diff(f, x, y)}")
print(f"∂²f/∂y∂x = {sp.diff(f, y, x)}")
print(f"혼합 편미분 교환: {sp.simplify(sp.diff(f, x, y) - sp.diff(f, y, x)) == 0}")

1.2 Total Differential¶

The total differential of a function $f(x, y, z)$:

$$ df = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz $$

This is a first-order approximation of the change in $f$ due to independent changes in each variable.

Physical meaning: The infinitesimal change in pressure $P(V, T)$ is: $$ dP = \left(\frac{\partial P}{\partial V}\right)_T dV + \left(\frac{\partial P}{\partial T}\right)_V dT $$

The subscript indicates the variable being held constant. This notation is very important in thermodynamics.

# 이상기체 PV = nRT에서의 전미분
P, V, T, n, R = sp.symbols('P V T n R', positive=True)

# P = nRT/V
P_expr = n * R * T / V

dP_dV = sp.diff(P_expr, V)
dP_dT = sp.diff(P_expr, T)

print(f"P = {P_expr}")
print(f"(∂P/∂V)_T = {dP_dV}")
print(f"(∂P/∂T)_V = {dP_dT}")
print(f"dP = ({dP_dV})dV + ({dP_dT})dT")

1.3 Gradient¶

The gradient of a scalar function $f(x, y, z)$ is a vector whose components are the partial derivatives:

$$ \nabla f = \left(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right) $$

Meaning of the gradient: - Direction of $\nabla f$: direction of steepest increase of $f$ - $|\nabla f|$: rate of increase in that direction (maximum directional derivative) - $\nabla f$ is perpendicular to the level surface $f = \text{const}$

import matplotlib.pyplot as plt

# 2D 기울기 시각화
f_2d = x**2 + y**2  # f(x,y) = x² + y²
grad_f = [sp.diff(f_2d, x), sp.diff(f_2d, y)]
print(f"f = {f_2d}")
print(f"∇f = ({grad_f[0]}, {grad_f[1]})")

# 수치 시각화
X, Y = np.meshgrid(np.linspace(-2, 2, 20), np.linspace(-2, 2, 20))
Z = X**2 + Y**2
U = 2 * X  # ∂f/∂x
V_field = 2 * Y  # ∂f/∂y

fig, ax = plt.subplots(figsize=(8, 7))
contour = ax.contour(X, Y, Z, levels=10, cmap='viridis')
ax.clabel(contour, inline=True, fontsize=8)
ax.quiver(X, Y, U, V_field, color='red', alpha=0.6, scale=50)
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_title('등고선과 기울기 벡터 (∇f ⊥ 등고선)')
ax.set_aspect('equal')
plt.tight_layout()
plt.savefig('gradient_field.png', dpi=100, bbox_inches='tight')
plt.close()

2. Chain Rule¶

2.1 Multivariable Chain Rule¶

If $f(x, y)$ where $x = x(t)$, $y = y(t)$:

$$ \frac{df}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} $$

More generally, if $f(x, y)$ where $x = x(s, t)$, $y = y(s, t)$:

$$ \frac{\partial f}{\partial s} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial s} $$

$$ \frac{\partial f}{\partial t} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial t} $$

Matrix notation (Jacobian):

$$ \begin{pmatrix} \frac{\partial f}{\partial s} \\ \frac{\partial f}{\partial t} \end{pmatrix} = \begin{pmatrix} \frac{\partial x}{\partial s} & \frac{\partial y}{\partial s} \\ \frac{\partial x}{\partial t} & \frac{\partial y}{\partial t} \end{pmatrix} \begin{pmatrix} \frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \end{pmatrix} $$

# 연쇄법칙 예제: 극좌표 변환
r, theta = sp.symbols('r theta', positive=True)
x_expr = r * sp.cos(theta)
y_expr = r * sp.sin(theta)

# f(x,y) = x² + y² 를 극좌표로 표현
f_xy = x**2 + y**2
f_polar = f_xy.subs([(x, x_expr), (y, y_expr)])
f_polar_simplified = sp.simplify(f_polar)
print(f"f(x,y) = {f_xy}")
print(f"f(r,θ) = {f_polar_simplified}")  # r²

# 연쇄법칙으로 ∂f/∂r 계산
df_dr_chain = sp.diff(f_xy, x).subs(x, x_expr).subs(y, y_expr) * sp.diff(x_expr, r) + \
              sp.diff(f_xy, y).subs(x, x_expr).subs(y, y_expr) * sp.diff(y_expr, r)
df_dr_direct = sp.diff(f_polar_simplified, r)

print(f"\n연쇄법칙: ∂f/∂r = {sp.simplify(df_dr_chain)}")
print(f"직접 미분: ∂f/∂r = {df_dr_direct}")

# 야코비안
J = sp.Matrix([[sp.diff(x_expr, r), sp.diff(x_expr, theta)],
               [sp.diff(y_expr, r), sp.diff(y_expr, theta)]])
print(f"\n야코비안 J =\n{J}")
print(f"det(J) = {sp.simplify(J.det())}")  # r (극좌표 야코비안)

2.2 Implicit Differentiation¶

When $y$ is implicitly defined as a function of $x$ by $F(x, y) = 0$:

$$ \frac{dy}{dx} = -\frac{\partial F / \partial x}{\partial F / \partial y} = -\frac{F_x}{F_y} $$

Derivation: Differentiating both sides of $F(x, y(x)) = 0$ with respect to $x$:

$$ \frac{\partial F}{\partial x} + \frac{\partial F}{\partial y}\frac{dy}{dx} = 0 $$

Extension to three variables: If $z = z(x, y)$ from $F(x, y, z) = 0$:

$$ \frac{\partial z}{\partial x} = -\frac{F_x}{F_z}, \qquad \frac{\partial z}{\partial y} = -\frac{F_y}{F_z} $$

# 음함수 미분 예제: 타원 x²/4 + y²/9 = 1
F = x**2/4 + y**2/9 - 1

dy_dx = -sp.diff(F, x) / sp.diff(F, y)
print(f"F(x,y) = {F} = 0")
print(f"dy/dx = {dy_dx}")
print(f"      = {sp.simplify(dy_dx)}")

# 검증: y = 3√(1 - x²/4) 를 직접 미분
y_explicit = 3 * sp.sqrt(1 - x**2/4)
dy_dx_explicit = sp.diff(y_explicit, x)
print(f"\n명시적 미분: dy/dx = {dy_dx_explicit}")
print(f"간소화: {sp.simplify(dy_dx_explicit)}")

# 순환 관계 (cyclic relation)
# F(x,y,z) = 0일 때:
# (∂x/∂y)_z (∂y/∂z)_x (∂z/∂x)_y = -1
print("\n=== 순환 관계 (이상기체) ===")
# PV = nRT → F(P,V,T) = PV - nRT = 0
P_sym, V_sym, T_sym = sp.symbols('P V T', positive=True)
F_gas = P_sym * V_sym - n * R * T_sym

dP_dV_T = -sp.diff(F_gas, V_sym) / sp.diff(F_gas, P_sym)  # (∂P/∂V)_T
dV_dT_P = -sp.diff(F_gas, T_sym) / sp.diff(F_gas, V_sym)  # (∂V/∂T)_P
dT_dP_V = -sp.diff(F_gas, P_sym) / sp.diff(F_gas, T_sym)  # (∂T/∂P)_V

product = sp.simplify(dP_dV_T * dV_dT_P * dT_dP_V)
print(f"(∂P/∂V)_T = {dP_dV_T}")
print(f"(∂V/∂T)_P = {dV_dT_P}")
print(f"(∂T/∂P)_V = {dT_dP_V}")
print(f"곱 = {product}")  # -1

3. Extrema and Saddle Points¶

3.1 Necessary Condition for Extrema¶

A critical point of a multivariable function $f(x, y)$: $\nabla f = 0$

$$ \frac{\partial f}{\partial x} = 0, \quad \frac{\partial f}{\partial y} = 0 $$

3.2 Second Derivative Test¶

At a critical point $(x_0, y_0)$, the Hessian matrix:

$$ H = \begin{pmatrix} f_{xx} & f_{xy} \\ f_{yx} & f_{yy} \end{pmatrix} $$

Discriminant:

$$ D = f_{xx} f_{yy} - (f_{xy})^2 = \det(H) $$

Condition	Conclusion
$D > 0$, $f_{xx} > 0$	Local minimum
$D > 0$, $f_{xx} < 0$	Local maximum
$D < 0$	Saddle point
$D = 0$	Test inconclusive (higher order needed)

Connection to positive definiteness: $D > 0$ and $f_{xx} > 0$ is equivalent to the Hessian $H$ being positive definite. This means the quadratic form $\frac{1}{2}\delta\mathbf{x}^T H \delta\mathbf{x} > 0$, hence a minimum.

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D

# 안장점을 포함하는 함수
f_saddle = x**2 - y**2  # 쌍곡 포물면

# 임계점 찾기
fx = sp.diff(f_saddle, x)
fy = sp.diff(f_saddle, y)
critical = sp.solve([fx, fy], [x, y])
print(f"f(x,y) = {f_saddle}")
print(f"임계점: {critical}")

# 헤시안
fxx = sp.diff(f_saddle, x, 2)
fxy = sp.diff(f_saddle, x, y)
fyy = sp.diff(f_saddle, y, 2)
D = fxx * fyy - fxy**2
print(f"\nH = [[{fxx}, {fxy}], [{fxy}, {fyy}]]")
print(f"D = {D}")
print(f"D < 0 → 안장점")

# 다른 예: 극소, 극대, 안장점을 모두 가진 함수
g = x**3 - 3*x*y**2  # 원숭이 안장 (monkey saddle)
h = (x**2 + y**2)**2 - 2*(x**2 - y**2)  # 두 극소, 두 안장점

# 3D 시각화
X_grid = np.linspace(-2, 2, 100)
Y_grid = np.linspace(-2, 2, 100)
X_mesh, Y_mesh = np.meshgrid(X_grid, Y_grid)

fig = plt.figure(figsize=(15, 5))

# 극소: x² + y²
ax1 = fig.add_subplot(131, projection='3d')
Z1 = X_mesh**2 + Y_mesh**2
ax1.plot_surface(X_mesh, Y_mesh, Z1, cmap='viridis', alpha=0.7)
ax1.set_title('극소: f = x² + y²')
ax1.set_xlabel('x')
ax1.set_ylabel('y')

# 안장점: x² - y²
ax2 = fig.add_subplot(132, projection='3d')
Z2 = X_mesh**2 - Y_mesh**2
ax2.plot_surface(X_mesh, Y_mesh, Z2, cmap='RdBu', alpha=0.7)
ax2.set_title('안장점: f = x² - y²')
ax2.set_xlabel('x')
ax2.set_ylabel('y')

# 극대: -(x² + y²)
ax3 = fig.add_subplot(133, projection='3d')
Z3 = -(X_mesh**2 + Y_mesh**2)
ax3.plot_surface(X_mesh, Y_mesh, Z3, cmap='plasma', alpha=0.7)
ax3.set_title('극대: f = -(x² + y²)')
ax3.set_xlabel('x')
ax3.set_ylabel('y')

plt.tight_layout()
plt.savefig('critical_points.png', dpi=100, bbox_inches='tight')
plt.close()

3.3 Practical Example: Distance Minimization¶

Problem of minimizing the distance from point $(x, y, z)$ to plane $2x + y - z = 5$.

Method 1: Minimize the squared distance $d^2 = (x-a)^2 + (y-b)^2 + (z-c)^2$ (covered in the next section on Lagrange multipliers)

Method 2: Use the distance formula directly $d = |2x + y - z - 5| / \sqrt{4 + 1 + 1}$

4. Lagrange Multipliers¶

4.1 Optimization with Equality Constraints¶

Problem: Find the extrema of $f(x, y)$ subject to the constraint $g(x, y) = 0$.

Lagrange function:

$$ \mathcal{L}(x, y, \lambda) = f(x, y) - \lambda \, g(x, y) $$

Necessary conditions:

$$ \frac{\partial \mathcal{L}}{\partial x} = 0, \quad \frac{\partial \mathcal{L}}{\partial y} = 0, \quad \frac{\partial \mathcal{L}}{\partial \lambda} = 0 $$

The third condition is the constraint $g = 0$ itself.

Geometric meaning: At an extremum, $\nabla f$ and $\nabla g$ are parallel, i.e., $\nabla f = \lambda \nabla g$.

# 예제: 원 x² + y² = 1 위에서 f(x,y) = x + y의 최댓값
lam = sp.Symbol('lambda')

f_obj = x + y
g_constraint = x**2 + y**2 - 1

L = f_obj - lam * g_constraint

# 연립방정식
eqs = [sp.diff(L, x), sp.diff(L, y), sp.diff(L, lam)]
print("라그랑주 조건:")
for eq in eqs:
    print(f"  {eq} = 0")

solutions = sp.solve(eqs, [x, y, lam])
print(f"\n해: {solutions}")

for sol in solutions:
    val = f_obj.subs([(x, sol[0]), (y, sol[1])])
    print(f"  ({sol[0]}, {sol[1]}): f = {val}, λ = {sol[2]}")

4.2 Multiple Constraints¶

With $m$ constraints $g_1 = 0, g_2 = 0, \ldots, g_m = 0$:

$$ \mathcal{L} = f - \lambda_1 g_1 - \lambda_2 g_2 - \cdots - \lambda_m g_m $$

Physics Application: Entropy Maximization

In statistical mechanics, the Boltzmann distribution is the solution to the following optimization problem:

Maximize: Entropy $S = -k_B \sum p_i \ln p_i$
Constraint 1: $\sum p_i = 1$ (normalization of probabilities)
Constraint 2: $\sum p_i E_i = \langle E \rangle$ (fixed average energy)

# 라그랑주 승수법으로 볼츠만 분포 유도 (이산 경우)
# 3개 에너지 준위 E₁=0, E₂=1, E₃=2
p1, p2, p3 = sp.symbols('p1 p2 p3', positive=True)
lam1, lam2 = sp.symbols('lambda1 lambda2')

E_levels = [0, 1, 2]
probs = [p1, p2, p3]

# 엔트로피 (부호 반전하여 최소화로 변환)
S = sum(p * sp.ln(p) for p in probs)  # 최소화할 것 (-S)

# 구속 조건
g1 = p1 + p2 + p3 - 1
g2 = 0*p1 + 1*p2 + 2*p3 - sp.Symbol('E_avg')

L = S + lam1 * g1 + lam2 * g2

eqs = [sp.diff(L, p) for p in probs] + [g1]
print("라그랑주 조건 (∂L/∂pᵢ = 0):")
for i, eq in enumerate(eqs[:3]):
    print(f"  ln(p{i+1}) + 1 + λ₁ + {E_levels[i]}λ₂ = 0")

print("\n→ pᵢ ∝ exp(-λ₂ Eᵢ) : 볼츠만 분포!")

4.3 Numerical Optimization with SciPy¶

from scipy.optimize import minimize

# 타원 x²/4 + y²/9 = 1 위에서 원점으로부터 최대/최소 거리
def objective(xy):
    return -(xy[0]**2 + xy[1]**2)  # 최대화 → 부호 반전

def constraint(xy):
    return xy[0]**2/4 + xy[1]**2/9 - 1  # = 0

from scipy.optimize import minimize
result = minimize(objective, x0=[1, 1],
                  constraints={'type': 'eq', 'fun': constraint},
                  method='SLSQP')

print(f"타원 위 원점에서 가장 먼 점: ({result.x[0]:.4f}, {result.x[1]:.4f})")
print(f"최대 거리: {np.sqrt(-result.fun):.4f}")

# 가장 가까운 점
result_min = minimize(lambda xy: xy[0]**2 + xy[1]**2, x0=[1, 0.1],
                       constraints={'type': 'eq', 'fun': constraint},
                       method='SLSQP')
print(f"타원 위 원점에서 가장 가까운 점: ({result_min.x[0]:.4f}, {result_min.x[1]:.4f})")
print(f"최소 거리: {np.sqrt(result_min.fun):.4f}")

5. Exact Differentials and Thermodynamics¶

5.1 Condition for Exact Differentials¶

For a differential form $M(x,y)\,dx + N(x,y)\,dy$ to be an exact differential, i.e., for there to exist a function $f$ such that $df = M\,dx + N\,dy$:

$$ \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} $$

This is a direct consequence of Schwarz's theorem: If $M = \partial f / \partial x$, $N = \partial f / \partial y$, then $\partial M / \partial y = \partial^2 f / \partial y \partial x = \partial^2 f / \partial x \partial y = \partial N / \partial x$.

# 완전미분 판별 예제
# (2xy + 3)dx + (x² + 4y)dy
M = 2*x*y + 3
N = x**2 + 4*y

dM_dy = sp.diff(M, y)
dN_dx = sp.diff(N, x)
print(f"M = {M}, N = {N}")
print(f"∂M/∂y = {dM_dy}, ∂N/∂x = {dN_dx}")
print(f"완전미분: {dM_dy == dN_dx}")

# 포텐셜 함수 f 구하기
# f = ∫M dx = x²y + 3x + h(y)
# ∂f/∂y = x² + h'(y) = N = x² + 4y
# → h'(y) = 4y → h(y) = 2y²
f_potential = x**2*y + 3*x + 2*y**2
print(f"\n포텐셜 함수: f = {f_potential}")
print(f"검증: ∂f/∂x = {sp.diff(f_potential, x)} (= M)")
print(f"검증: ∂f/∂y = {sp.diff(f_potential, y)} (= N)")

5.2 Thermodynamics and Maxwell Relations¶

Combining the first and second laws of thermodynamics:

$$ dU = TdS - PdV $$

Since this is an exact differential ($U = U(S, V)$):

$$ T = \left(\frac{\partial U}{\partial S}\right)_V, \quad -P = \left(\frac{\partial U}{\partial V}\right)_S $$

From the condition for exact differentials $\partial^2 U / \partial V \partial S = \partial^2 U / \partial S \partial V$:

$$ \left(\frac{\partial T}{\partial V}\right)_S = -\left(\frac{\partial P}{\partial S}\right)_V $$

This is one of the Maxwell relations.

Four thermodynamic potentials and Maxwell relations:

Potential	Differential	Maxwell Relation
$U(S,V)$	$dU = TdS - PdV$	$\left(\frac{\partial T}{\partial V}\right)_S = -\left(\frac{\partial P}{\partial S}\right)_V$
$H(S,P)$	$dH = TdS + VdP$	$\left(\frac{\partial T}{\partial P}\right)_S = \left(\frac{\partial V}{\partial S}\right)_P$
$F(T,V)$	$dF = -SdT - PdV$	$\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V$
$G(T,P)$	$dG = -SdT + VdP$	$-\left(\frac{\partial S}{\partial P}\right)_T = \left(\frac{\partial V}{\partial T}\right)_P$

Mnemonic (Born square): By placing internal energy at the vertices and natural variables at the edges, you can systematically derive Maxwell relations.

# 맥스웰 관계식 검증 (이상기체)
# PV = nRT → P = nRT/V
S, V_td = sp.symbols('S V', positive=True)
k, N_a = sp.symbols('k N_A', positive=True)

# 이상기체의 내부 에너지: U = (3/2)nRT = (3/2)nR * T(S,V)
# 간단한 모델: U(S,V) = A * exp(2S/3nR) * V^(-2/3)
# 여기서는 맥스웰 관계식의 구조만 확인

# 헬름홀츠 자유에너지 F(T,V)에서
# dF = -SdT - PdV
# 이상기체: F = nRT(1 - ln(T/T₀)) - nRT ln(V/V₀) + const

T_td = sp.Symbol('T', positive=True)
F_ideal = n*R*T_td*(1 - sp.ln(T_td)) - n*R*T_td*sp.ln(V_td)

S_from_F = -sp.diff(F_ideal, T_td)
P_from_F = -sp.diff(F_ideal, V_td)

print(f"F = {F_ideal}")
print(f"S = -∂F/∂T = {sp.simplify(S_from_F)}")
print(f"P = -∂F/∂V = {sp.simplify(P_from_F)}")

# 맥스웰: (∂S/∂V)_T = (∂P/∂T)_V
dS_dV = sp.diff(S_from_F, V_td)
dP_dT = sp.diff(P_from_F, T_td)
print(f"\n(∂S/∂V)_T = {dS_dV}")
print(f"(∂P/∂T)_V = {dP_dT}")
print(f"맥스웰 관계 성립: {sp.simplify(dS_dV - dP_dT) == 0}")

6. Multivariate Taylor Series¶

6.1 Two-Variable Taylor Expansion¶

Expansion of $f(x, y)$ near the point $(a, b)$:

$$ f(x, y) = f(a, b) + f_x(a,b)(x-a) + f_y(a,b)(y-b) \\ + \frac{1}{2!}\left[f_{xx}(x-a)^2 + 2f_{xy}(x-a)(y-b) + f_{yy}(y-b)^2\right] + \cdots $$

Vector notation: $\mathbf{x}_0 = (a, b)$, $\delta\mathbf{x} = (x-a, y-b)$

$$ f(\mathbf{x}_0 + \delta\mathbf{x}) = f(\mathbf{x}_0) + \nabla f \cdot \delta\mathbf{x} + \frac{1}{2} \delta\mathbf{x}^T H \, \delta\mathbf{x} + \cdots $$

where $H$ is the Hessian matrix.

6.2 Physics Approximations¶

Small-amplitude approximation: Potential energy of a pendulum $U(\theta) = mgl(1 - \cos\theta)$ near $\theta = 0$:

$$ U \approx mgl \cdot \frac{\theta^2}{2} + O(\theta^4) \quad (\text{harmonic oscillator}) $$

Multivariable function approximation: $f(x, y) = e^{x}\sin(y)$ near the origin:

$$ f \approx y + xy + \frac{1}{2}(y - \frac{y^3}{6} + \cdots) \approx y + xy - \frac{y^3}{6} + \cdots $$

# 다변수 테일러 급수
f_taylor = sp.exp(x) * sp.sin(y)

# 원점에서 3차까지 전개
taylor_2d = sp.series(sp.series(f_taylor, x, 0, 4), y, 0, 4)
print(f"f = exp(x)sin(y)")
print(f"테일러 (3차): {taylor_2d}")

# SymPy의 다변수 테일러
from sympy import O
# 수동 전개
f0 = f_taylor.subs([(x, 0), (y, 0)])
fx0 = sp.diff(f_taylor, x).subs([(x, 0), (y, 0)])
fy0 = sp.diff(f_taylor, y).subs([(x, 0), (y, 0)])
fxx0 = sp.diff(f_taylor, x, 2).subs([(x, 0), (y, 0)])
fxy0 = sp.diff(f_taylor, x, y).subs([(x, 0), (y, 0)])
fyy0 = sp.diff(f_taylor, y, 2).subs([(x, 0), (y, 0)])

T2 = f0 + fx0*x + fy0*y + sp.Rational(1,2)*(fxx0*x**2 + 2*fxy0*x*y + fyy0*y**2)
print(f"\n2차 테일러: {T2}")

# 근사 정확도 비교
import matplotlib.pyplot as plt

x_vals = np.linspace(-1, 1, 50)
y_vals = np.linspace(-np.pi/2, np.pi/2, 50)
X_mesh, Y_mesh = np.meshgrid(x_vals, y_vals)

Z_exact = np.exp(X_mesh) * np.sin(Y_mesh)
Z_approx = Y_mesh + X_mesh*Y_mesh + 0.5*(X_mesh**2*Y_mesh)  # 주요 항

fig, axes = plt.subplots(1, 3, figsize=(15, 4))
for ax, Z, title in [(axes[0], Z_exact, '정확한 값'),
                       (axes[1], Z_approx, '2차 근사'),
                       (axes[2], Z_exact - Z_approx, '오차')]:
    c = ax.contourf(X_mesh, Y_mesh, Z, levels=20, cmap='RdBu_r')
    plt.colorbar(c, ax=ax)
    ax.set_xlabel('x')
    ax.set_ylabel('y')
    ax.set_title(title)

plt.tight_layout()
plt.savefig('taylor_2d.png', dpi=100, bbox_inches='tight')
plt.close()

7. Change of Variables and Jacobian¶

7.1 Jacobian Matrix and Determinant¶

The Jacobian matrix of a coordinate transformation $(x, y) \to (u, v)$:

$$ J = \frac{\partial(x, y)}{\partial(u, v)} = \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix} $$

Jacobian (determinant): $\left|\frac{\partial(x, y)}{\partial(u, v)}\right| = \det(J)$

Change of variables in multiple integrals:

$$ \iint f(x, y) \, dx \, dy = \iint f(x(u,v), y(u,v)) \left|\frac{\partial(x, y)}{\partial(u, v)}\right| du \, dv $$

7.2 Jacobians of Common Coordinate Transformations¶

Transformation	Jacobian
Polar $(r, \theta)$	$r$
Cylindrical $(r, \theta, z)$	$r$
Spherical $(r, \theta, \phi)$	$r^2 \sin\theta$

# 야코비안 계산 예제
r, theta, phi = sp.symbols('r theta phi', positive=True)

# 구면좌표 → 직교좌표
x_sph = r * sp.sin(theta) * sp.cos(phi)
y_sph = r * sp.sin(theta) * sp.sin(phi)
z_sph = r * sp.cos(theta)

J_spherical = sp.Matrix([
    [sp.diff(x_sph, r), sp.diff(x_sph, theta), sp.diff(x_sph, phi)],
    [sp.diff(y_sph, r), sp.diff(y_sph, theta), sp.diff(y_sph, phi)],
    [sp.diff(z_sph, r), sp.diff(z_sph, theta), sp.diff(z_sph, phi)]
])

jacobian_det = sp.simplify(J_spherical.det())
print("구면좌표 야코비안 행렬:")
sp.pprint(J_spherical)
print(f"\n|J| = {jacobian_det}")  # r² sin(θ)

# 체적 요소
print(f"\ndV = |J| dr dθ dφ = {jacobian_det} dr dθ dφ")

# 응용: 구의 체적
# V = ∫₀^a ∫₀^π ∫₀^{2π} r² sin(θ) dφ dθ dr
a = sp.Symbol('a', positive=True)
volume = sp.integrate(
    sp.integrate(
        sp.integrate(r**2 * sp.sin(theta), (phi, 0, 2*sp.pi)),
        (theta, 0, sp.pi)),
    (r, 0, a))
print(f"\n구의 체적 = {volume} = {sp.simplify(volume)}")  # (4/3)πa³

Exercises¶

Basic Problems¶

1. Find all first and second partial derivatives of $f(x, y) = x^3 y^2 + \sin(xy)$ and verify that $\frac{\partial^2 f}{\partial x \partial y} = \frac{\partial^2 f}{\partial y \partial x}$.

2. Express the Laplacian $\nabla^2 f = \frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2}$ in polar coordinates $(r, \theta)$ in terms of $r$ and $\theta$.

3. Find all critical points of $f(x, y) = x^3 + y^3 - 3xy$ and classify each type (maximum/minimum/saddle point).

Applied Problems¶

4. Using the Lagrange multiplier method, find the maximum and minimum values of $f(x, y, z) = x + 2y + 3z$ on the sphere $x^2 + y^2 + z^2 = 1$.

5. For the van der Waals equation $(P + a/V^2)(V - b) = RT$, find $\left(\frac{\partial P}{\partial T}\right)_V$ and $\left(\frac{\partial V}{\partial T}\right)_P$.

6. Calculate the three-dimensional Gaussian integral $\iiint e^{-(ax^2 + by^2 + cz^2)} \, dx \, dy \, dz$ (integration domain: all space) using spherical coordinate transformation and the Jacobian. (Hint: separate after principal axis transformation)

References¶

Boas, Mathematical Methods in the Physical Sciences, Chapter 4
Arfken & Weber, Mathematical Methods for Physicists, Chapter 1
Callen, Thermodynamics and an Introduction to Thermostatistics (Maxwell relations)
Previous lesson: 03. Linear Algebra (matrices, quadratic forms)
Next lesson: 05. Vector Analysis

Next Lesson¶

05. Vector Analysis covers gradient, divergence, curl operators and integral theorems in depth.